The size of proper classes?
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From: Aatu Koskensilta on 4 Dec 2009 15:20 zuhair <zaljohar(a)gmail.com> writes: > Well the problem against that is that it is not a theorem of ZF that > every set is the same size of a well founded set. The problem against what? I was replying to your comment that you "got they idea that they don't work outside Regularity" about Scott cardinals, and noted that this is indeed so, unless we adopt some additional axiom. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: zuhair on 4 Dec 2009 15:23 On Dec 4, 12:45 pm, hru...(a)odds.stat.purdue.edu (Herman Rubin) wrote: > In article <09620f7e-a9c0-442a-a476-ee5114e4c...(a)p19g2000vbq.googlegroups..com>, > George Greene <gree...(a)email.unc.edu> wrote: > > >On Dec 3, 12:16=A0pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > >> hru...(a)odds.stat.purdue.edu (Herman Rubin) writes: > >> > I am almost certain that there exist models with proper classes > >> > strictly larger than the class of all ordinal numbers, and all > >> > comparable. > >So the class of all ordinal numbers, in this context, is proper, but > >is comparably smaller than larger proper classes? > > Possibly, and possibly not. If V = L, as in Godel's model > which proves the consistency of the generalized continuum > hypothesis, the answer is not; the entire universe has the > same "cardinality" as the class of all ordinal numbers. > > >And in this context, there can be ordinal proper classes as well? > > In NBG, there must be. > > >And the proper class of all set ordinals is one of them? > > Yes. > > >And this proper-class ordinal also has proper-class successors that > >are also ordinals? > > Yes. How is that possible? a proper class is a class that is not a member of any class, on the other hand the definition of ordinal successor of the ordinal x for example is x U {x}, i.e. if y is the ordinal successor of the ordinal x, then x must be a member of y, now if x is a proper class ordinal, then it cannot have a successor, otherwise x would be a proper class that is a member of a class, which is contradictive.That is the case of course in NBG\MK. However in Ackermann's class theory, every ordinal weather it is a set or a proper class, must have a successor, since Ackermann's theory becomes inconsistent if there is no ordinal successor for an ordinal. But in Ackermann's proper classes are not defined as not being members of other classes, they are defined as not being sets. So I think the above reply made by Herman Rubin is mistaken, if we are working in NBG\MK. > > Do these proper-class ordinals eventually get > > >big enough to be equipollent to the largest proper classes? > > Every constructible ordinal class has the same cardinality as > the class of ordinal numbers. One cannot talk about the class > of these in NBG, but might be able to in Morse-Kelley. The > largest proper class is the universe in any case, and the > same question remains. > -- > This address is for information only. I do not claim that these views > are those of the Statistics Department or of Purdue University. > Herman Rubin, Department of Statistics, Purdue University > hru...(a)stat.purdue.edu Phone: (765)494-6054 FAX: (765)494-0558
From: Aatu Koskensilta on 4 Dec 2009 15:24 zuhair <zaljohar(a)gmail.com> writes: > How is that possible? a proper class is a class that is not a member > of any class, on the other hand the definition of ordinal successor of > the ordinal x for example is x U {x}, i.e. if y is the ordinal > successor of the ordinal x, then x must be a member of y, now if x is > a proper class ordinal, then it cannot have a successor, that is of > course in NBG\MK. Presumably Herman had in mind class sized well-orderings of order type greater than On, the class of ordinals. In NBG and MK we can talk of On + 1, On * 2, and so on, understood in this sense. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: zuhair on 4 Dec 2009 15:34 On Dec 4, 3:20 pm, Aatu Koskensilta <aatu.koskensi...(a)uta.fi> wrote: > zuhair <zaljo...(a)gmail.com> writes: > > Well the problem against that is that it is not a theorem of ZF that > > every set is the same size of a well founded set. > > The problem against what? I was replying to your comment that you "got > they idea that they don't work outside Regularity" about Scott > cardinals, and noted that this is indeed so, unless we adopt some > additional axiom. Yes, I agree with your agreement with me about "Regularity", but I want to make sure of one point though, does Scott cardinals work outside Regularity if we assume choice. my guess is that they don't, because a Von Neumann cardinal is not necessarily of the least rank, does that make sense? However the problem that I was saying is regarding having well founded cardinals for all sets in the universe, I was speaking somewhat of a more general matter, were even if we amend the definition of Scott cardinals as you did, yet this amendment that you've done will not be enough to make those well founded Scott cardinals defined for every set of the universe in the desirable manner that we want from cardinals, unless we assume Jean Coret's assumption that every set is equinumerous to some well founded set, then we can have empty cardinals for non equinumerous sets, thus departing from the desirable quality we want cardinals to fulfill. To speak more precisely your amendment will actually define cardinals for every set, and those cardinal will be sets always!, but the problem in having them non empty. > > -- > Aatu Koskensilta (aatu.koskensi...(a)uta.fi) > > "Wovon man nicht sprechan kann, darüber muss man schweigen" > - Ludwig Wittgenstein, Tractatus Logico-Philosophicus
From: Aatu Koskensilta on 4 Dec 2009 15:40
zuhair <zaljohar(a)gmail.com> writes: > Yes, I agree with your agreement with me about "Regularity", but I > want to make sure of one point though, does Scott cardinals work > outside Regularity if we assume choice. my guess is that they don't, > because a Von Neumann cardinal is not necessarily of the least rank, > does that make sense? The rank of a von Neumann ordinal alpha is alpha. I'm not sure exactly what you're asking, but Scott cardinals certainly work when we have choice, since given choice every set is the same size as an ordinal, and all ordinals are well-founded. If we have choice, it is more natural to define the cardinal of a set to be the initial ordinal equipollent to it. > To speak more precisely your amendment will actually define cardinals > for every set, and those cardinal will be sets always!, but the > problem in having them non empty. Yes, I explicitly noted this in my post: unless we assume (or prove) that every set is the same size as a well-founded set we can't use Scott's trick to define cardinals. -- Aatu Koskensilta (aatu.koskensilta(a)uta.fi) "Wovon man nicht sprechan kann, dar�ber muss man schweigen" - Ludwig Wittgenstein, Tractatus Logico-Philosophicus |